Question: User: $\frac{\sqrt{10}+\sqrt{10}+\sqrt{10}}{\sqrt{30}}$ = $81^{x^{3}}$
1. **State the problem:**
Solve for $x$ in the equation $$\frac{\sqrt{10}+\sqrt{10}+\sqrt{10}}{\sqrt{30}} = 81^{x^{3}}.$$
2. **Simplify the left side:**
Since $\sqrt{10}$ is repeated three times in the numerator, we have:
$$\frac{\sqrt{10}+\sqrt{10}+\sqrt{10}}{\sqrt{30}} = \frac{3\sqrt{10}}{\sqrt{30}}.$$
3. **Simplify the fraction:**
Rewrite the denominator:
$$\sqrt{30} = \sqrt{3 \times 10} = \sqrt{3} \times \sqrt{10}.$$
So the fraction becomes:
$$\frac{3\sqrt{10}}{\sqrt{3} \times \sqrt{10}}.$$
Cancel $\sqrt{10}$ from numerator and denominator:
$$\frac{3\cancel{\sqrt{10}}}{\sqrt{3} \times \cancel{\sqrt{10}}} = \frac{3}{\sqrt{3}}.$$
4. **Rationalize the denominator:**
Multiply numerator and denominator by $\sqrt{3}$:
$$\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3}.$$
Cancel 3:
$$\frac{\cancel{3}\sqrt{3}}{\cancel{3}} = \sqrt{3}.$$
So the left side simplifies to $\sqrt{3}$.
5. **Rewrite the right side:**
Recall that $81 = 3^{4}$, so:
$$81^{x^{3}} = (3^{4})^{x^{3}} = 3^{4x^{3}}.$$
6. **Set the equation:**
$$\sqrt{3} = 3^{4x^{3}}.$$
Note that $\sqrt{3} = 3^{\frac{1}{2}}$, so:
$$3^{\frac{1}{2}} = 3^{4x^{3}}.$$
7. **Equate exponents:**
Since the bases are the same and positive (3), the exponents must be equal:
$$\frac{1}{2} = 4x^{3}.$$
8. **Solve for $x^{3}$:**
$$x^{3} = \frac{1}{2 \times 4} = \frac{1}{8}.$$
9. **Solve for $x$:**
Take the cube root of both sides:
$$x = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}.$$
**Final answer:**
$$\boxed{\frac{1}{2}}.$$